# Prime satisfying a given condition

Let $p$ be a prime. If $\frac{p-1}{4}$ and $\frac{p+1}{2}$ are also primes then prove that $p=13$.

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It would help if you offered some explanation or motivation for this question. – Akhil Mathew Aug 5 '10 at 21:07
As far as I can tell it is just a contest-style problem. No motivation aside from curiosity, probably. (The thing we have to watch out for is people posting problems from, say, magazine contests which are still ongoing, but I hope this won't be an issue.) – Qiaochu Yuan Aug 6 '10 at 5:27
If you consider 1 as a prime, which most do not, then p=5 is also a solution. – Heath Hunnicutt Aug 6 '10 at 20:41

nice one. For (p-1)/4 to be odd, p must be of the form 8k+5, so (p-1)/4 is of the form 2k+1 and (p+1)/2 of the form 4k+3.
If k is = 0,1,2 (mod 3) then the three numbers are respectively congruent to
k=0(mod 3): 2,1,0 (mod 3)
k=1(mod 3): 1,0,1 (mod 3)
k=2(mod 3): 0,2,2 (mod 3)
This means that the only way all three of them are prime numbers is that one of them is 3. For k=0 we have 5,1,3 which is ruled out; for k=1 we have 13,3,7 which satisfies hypotheses; for k>1 all numbers are > 3. The only other case to be checked is (p-1)/4 = 2; in this case however p = 9, so this could not be a solution.

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The primes have product $(p^3 - p)/8$ which is divisible by $3$. So one prime = $3$. The rest is trivial.

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The cases $p=2,3$ can be trivially checked. I'll assume $p\ge 5$.

Note $\left( \displaystyle\frac{p-1}{4} \right) \left( \displaystyle\frac{p+1}{2}\right)=\displaystyle\frac{p^2-1}{8}=a$ (say). Then $a$ has only two prime divisors.

Now it is well known that $24|p^2-1$ for $p>3$. Let $p^2-1=24t$. Then $a=3t$.

Thus $3$ is a prime divisor of $a$, implying $3$ equals one of $\displaystyle\frac{p-1}{4},\frac{p+1}{2}$. Direct substitution shows that $p=13$ is the only solution.

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